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Unformatted text preview: MATH 239 Assignment 5 Solutions Note: The notation e = uv is a shorthand for e = { u,v } . Even though we use the shorthand here, it is always understood that an edge is a subset of the vertices of size 2. 1. For each of the following descriptions of possible graphs, either show that one exists by drawing an example, or prove that such a graph does not exist. (a) A 3-regular graph that contains a bridge. (b) A 4-regular graph that contains a bridge. (c) A graph whose minimum degree is 4 (i.e. every vertex has degree at least 4) that contains a bridge. Solution. (a) Such a graph exists, for example, (b) No such graphs exist. Suppose that G is a 4-reguar graph with a bridge e = uv . Then G- e has two components, say H is the one containing u . Then every vertex in H has degree 4 except for u , which has degree 3 in H . So H is a graph that has exactly one vertex of odd degree, which contradicts the fact that every graph has even number of odd-degree vertices. Therefore, no 4-regular graphs contains a bridge. (c) Such a graph exists, for example, 2. Determine the minimum number of vertices r in any tree T having two vertices of degree 3, two vertices of degree 4, and one vertex of degree 6. Show that your answer is best possible by giving an example of such a tree with exactly r vertices. 1 Solution. Let p denote the number of vertices of T , and let n i denote the number of vertices of degree exactly i . Then p = n 1 + n 2 + n 3 + ··· where we know n 1 = 2 + n 3 + 2 n 4 + 3 n 5 + 4 n 6 + ··· . Therefore, p = (2 + n 3 + 2 n 4 + 3 n 5 + 4 n 6 + ··· ) + n 2 + n 3 + n 4 + ··· = 2 + n 2 + 2 n 3 + 3 n 4 + 4 n 5 + ··· ≥ 2 + 2 n 3 + 3 n 4 + 5 n 6 (since each n i is nonnegative) ≥ 2 + 2(2) + 3(2) + 5 = 17 ....
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